Thursday, June 13, 2013

Popper functions and null sets

Let's go back to the problem that I keep on thinking about: How to distinguish possibilities that are classically of null probability. For instance, given a uniform choice of a point on some nice set (say, a ball) in Euclidean space, we want to say something like P({x,y})>P({x}), when x and y are distinct: it's more likely that one would hit one of two points than that one would hit a particular point. A series of my blog posts (and at least one article) showed that infinitesimals aren't the way. What about conditional probabilities?

For various reasons, instead of taking unconditional probabilities to be fundamental and defining conditional probabilities in terms of them, one may want to take conditional probabilities as fundamental. The standard method is to use Popper functions (I'll assume the linked axioms below). One might hope to use Popper functions to do things like make sense of the difference between the probability of two points in a continuous case (say, where a point is uniformly chosen in some nice subset of a Euclidean space) and the probability of a single point. For instance, one might hope that P({x,y}|{x,y})>P({x}|{x,y}) whenever x and y are distinct.

This won't work, however. Instead of working with propositions, I will work with sets—the definitions of Popper functions neatly adapt. Let Ω be a solid three-dimensional ball. Assume that P(A|B) is defined for all A and B in some algebra F of subsets of Ω.

Say that a set A in F is P-trivial provided that P(C|A)=1 for all C (including for the empty set). The empty set is trivial, of course. In order for us to have any hope of saying things like P({x,y}|{x,y})>P({x}|{x,y}), we better have sets with two points be non-trivial. Now, it's not hard to show that a finite union of trivial sets is trivial and that the subsets of a trivial set are trivial, so in order for all the finite sets to be non-trivial, it's necessary and sufficient that singletons be non-trivial.

Moreover, we want rotational symmetry. Say that F and P are rotationally symmetric provided that for any rotation r around the origin and A and B in F, rA and rB are in F, and P(rA|rB)=P(A|B).

Theorem. If P is rotationally symmetric and F includes all countable subsets of Ω, then for any sphere of positive radius around the origin lying in Ω there is at least one P-trivial countably infinite set on the surface of
that sphere. Thus, all finite sets not containing the origin are trivial.

If we are to be extending something like classical probabilities, we do want countable subsets of Ω to be in F. The triviality of finite sets follows from the fact that some singleton on every sphere of non-zero radius about the origin is trivial (since any subset of a trivial set is trivial) and if one singleton on that sphere is trivial, then by rotational invariance, they all are, while finite unions of trivial sets are trivial. So all we need to prove is the existence of that trivial countably infinite set.

The proof is easily based on standard ideas from the proof of the Banach-Tarski Paradox.

Proof. Let SO(3) be the rotation group around the origin. Famously, there is a subgroup G that is isomorphic to the free group F2 on two elements. Choose a point ω in Ω such that ρ(ω)=ω only if ρ is the identity rotation. (This is a counting argument: G is a countable group, and each non-identity member of G has two fixed points on any given sphere around the center, so for any fixed sphere there will be only countably many fixed points of non-identity members of G on it, and hence there will be a point on the sphere that isn't a fixed point of any non-identity member.) Let H={ρ(ω):ρ∈G}. Then H is a countable subset of Ω.

For a reductio ad absurdum, suppose H is non-trivial. Then P(−|H) is a finitely additive probability measure on H. Moreover, H=ρH for any ρ∈G, so by rotational invariance P(ρK|H)=P(ρK|ρH)=P(K|H) for any ρ∈G, and so P(−|H) is a finitely additive G-invariant measure on H. Using the bijection between G and H given by f(ρ)=ρ(ω) (we use the choice of ω to see that f is one-to-one), we can then get a finitely additive G-left-invariant measure on G. But G is isomorphic to F2 and hence is not amenable and hence has no such invariant measure. (One could also neatly demonstrate a paradoxical decomposition here.) That's a contradiction, so H must be trivial. QED

Even though this argument uses ideas from the proof of Banach-Tarski, and famously the latter uses the Axiom of Choice, this argument does not use the Axiom of Choice.

One can perhaps get out of this by having a more onerous requirement on the sets A and B that are on either side of the bar in "P(A|B)" than that they are fit in a single algebra F. We want any countable set to be an acceptable A, but perhaps we don't want to allow every countable set as an acceptable B. I don't know what natural requirement could be put here.

5 comments:

Just replace H in my argument by a countable collection of equally-sized radial line segments at the same angular positions as the points in H were and with endpoints at, say, 1/4 and 1/2 units from the center.

1. If F includes all Lebesgue measurable sets, then all Lebesgue null sets are P-trivial.

So P can't be an *improvement* on Lebesgue measure, since for it to be such, it would have to measure all the Lebesgue measurable sets.

2. If F includes all countable sets, then all countable sets are P-trivial.

The proof is to take our Lebesgue null or countable (resp.) set A, find a subgroup G of SO(3) isomorphic to F_2 with the property that none of G's fixed points are in A. Let A' = GA. Then by mimicking the standard constructions in the Hausdorff / Banach-Tarski paradoxes, one obtains a G-paradoxical decomposition of A' (this uses Choice). Since G is countable, A' is also null or countable (resp.), and hence in F. Moreover, the G-paradoxical decomposition of A' is into subsets of A', and hence into sets that are null or countable (resp.), and hence in F, and so P(-|A') can't be a G-invariant measure on F. Hence A' must be P-trivial, and hence so must its subset A.

Actually, in the first case there might not be a subgroup of SO(3) without fixed points if we're talking of subsets of R^3, since the sphere has null 3-dimensional Lebesgue measure. Instead, I think one should just mimic parts of the proof of the improved Hausdorff paradox (improved by removing the exception set) in the Wagon book.

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I am a philosopher at Baylor University. This blog, however, does not purport to express in any way the opinions of Baylor University. Amateur science and technology work should not be taken to be approved by Baylor University. Use all information at your own risk.